\newproblem{lay:3_3_11}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 3.3.11}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Calculate the adjugate of the matrix $A=\begin{pmatrix}0 & -2 & -1\\3 & 0 & 0\\ -1 & 1 & 1\end{pmatrix}$. Then, use it to calculate $A^{-1}$.
}{
   % Solution
	For calculating the adjugate of the matrix $A$ we need to calculate all its cofactors
	\begin{center}
		$\begin{array}{rcccl}
			C_{11}&=&(-1)^{1+1}\left|\begin{array}{rr}0&0\\1&1\end{array}\right|&=&0\\
			C_{12}&=&(-1)^{1+2}\left|\begin{array}{rr}3&0\\-1&1\end{array}\right|&=&-3\\
			C_{13}&=&(-1)^{1+3}\left|\begin{array}{rr}3&0\\-1&1\end{array}\right|&=&3\\
			C_{21}&=&(-1)^{2+1}\left|\begin{array}{rr}-2&-1\\1&1\end{array}\right|&=&1\\
			C_{22}&=&(-1)^{2+2}\left|\begin{array}{rr}0&-1\\-1&1\end{array}\right|&=&-1\\
			C_{23}&=&(-1)^{2+3}\left|\begin{array}{rr}0&-2\\-1&1\end{array}\right|&=&2\\
			C_{31}&=&(-1)^{3+1}\left|\begin{array}{rr}-2&-1\\0&0\end{array}\right|&=&0\\
			C_{32}&=&(-1)^{3+2}\left|\begin{array}{rr}0&-1\\3&0\end{array}\right|&=&-3\\
			C_{33}&=&(-1)^{3+3}\left|\begin{array}{rr}0&-2\\3&0\end{array}\right|&=&6\\
		\end{array}$
	\end{center}
	The adjoint is
	\begin{center}
		$A^*=\left(\begin{array}{rrr}0 & -3 & 3 \\ 1 & -1 & 2 \\ 0 & -3 & 6 \end{array}\right)$
	\end{center}
	For calculating $A^{-1}$ we need the determinant of $A$. We use the cofactor expansion along the second row
	\begin{center}
		$|A|=a_{21}C_{21}+a_{22}C_{22}+a_{23}C_{23}=3\cdot 1=3$
	\end{center}
	Now
	\begin{center}
		$A^{-1}=\frac{1}{|A|}(A^*)^T=\frac{1}{3}\left(\begin{array}{rrr}0 & 1 & 0 \\ -3 & -1 & -3 \\ 3 & 2 & 6 \end{array}\right)=
		   \left(\begin{array}{rrr}0 & \frac{1}{3} & 0 \\ -1 & -\frac{1}{3} & -1 \\ 1 & \frac{2}{3} & 2 \end{array}\right)$
	\end{center}
}
\useproblem{lay:3_3_11}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
